ta có : số cần tìm có dạng \(\overline{ab7}\) (điều kiện \(a;b\in Z,1\le a\le3\))

vì nếu chuyển số 7 đó lên đầu thì ta đc 1 số mới mà khi chia cho số cũ thì đc thương 2 dư 21

\(\Rightarrow\dfrac{\overline{7ab}}{\overline{ab7}}=2+\dfrac{21}{\overline{ab7}}\)

th1: \(b=9\)

\(\Rightarrow\dfrac{\overline{7ab}}{\overline{ab7}}=2+\dfrac{21}{\overline{ab7}}=\dfrac{\overline{\left(2a+1\right)\left(8\right)4}+21}{\overline{ab7}}=\dfrac{\overline{\left(2a+2\right)\left(0\right)5}}{\overline{ab7}}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+2=7\\0=a\\5=b\end{matrix}\right.\left(L\right)\)

th2: \(5\le b< 9\)

\(\Rightarrow\dfrac{\overline{7ab}}{\overline{ab7}}=2+\dfrac{21}{\overline{ab7}}=\dfrac{\overline{\left(2a+1\right)\left(2b+1-10\right)4}+21}{\overline{ab7}}=\dfrac{\overline{\left(2a+1\right)\left(2b-7\right)5}}{\overline{ab7}}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+1=7\\2b-7=a\\5=b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=5\end{matrix}\right.\) (tmđk)

th3: \(b=4\)

\(\Rightarrow\dfrac{\overline{7ab}}{\overline{ab7}}=2+\dfrac{21}{\overline{ab7}}=\dfrac{\overline{\left(2a\right)\left(9\right)4}+21}{\overline{ab7}}=\dfrac{\overline{\left(2a+1\right)\left(1\right)5}}{\overline{ab7}}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+1=7\\1=a\\5=b\end{matrix}\right.\left(L\right)\)

th4: \(0\le b< 4\)

\(\Rightarrow\dfrac{\overline{7ab}}{\overline{ab7}}=2+\dfrac{21}{\overline{ab7}}=\dfrac{\overline{\left(2a\right)\left(2b+1\right)4}+21}{\overline{ab7}}=\dfrac{\overline{\left(2a\right)\left(2b+3\right)5}}{\overline{ab7}}\)

\(\Rightarrow\left\{{}\begin{matrix}2a=7\\2b+3=a\\5=b\end{matrix}\right.\left(L\right)\)

vậy số đó là \(357\)